Answer
$-\frac{1}{10}cos 5x+\frac{cos x}{2}+C$
Work Step by Step
Applying the product-to-sum identity, we get
$\int sin2x cos 3xdx=\frac{1}{2}\int[sin(-x)+sin5x]dx$
$=\frac{1}{2}\int(sin 5x- sinx)dx$
$= \frac{1}{2}\int sin 5xdx-\frac{1}{2}\int sin xdx$
$=(\frac{1}{2}\times\frac{-cos5x}{5})-(\frac{1}{2}\times-cos x)+C$
$= -\frac{1}{10}cos 5x+\frac{cos x}{2}+C$
